Physical Sciences Grade 12 P2 Mpumalanga November 2014 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE EXAM PHYSICAL SCIENCES P2 SEPTEMBER 2014 GRADE 12 MEMORANDUM MARKS: 150 TIME: 3 HOURS This memorandum consists of 10 pages

Downloaded from hlayiso.com Physical Sciences/P2 2 September 2014 NSC-Memorandum QUESTION 1 : MULTIPLE-CHOICE QUESTIONS 1.1 D (2) 1.2 B  (2) 1.3 B  (2) 1.4 A  (2) 1.5 A  (2) 1.6 D  (2) 1.7 B  (2) 1.8 D  (2) 1.9 C  (2) 1.10 A  (2) [20] QUESTION 2 2.1 Homologous series: A series of organic compounds that can be described by the same general formula with the same functional group and in which one member differs from the next with a -CH2- group.  (2) 2.2.1 H H H H H H H C C C C C C H H H H H H H H H H H (2) H C C C C C C H H H H H any one  Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P2 3 September 2014 NSC-Memorandum 2.2.2 H C O H  aldehydes / alkanals (2) 2.2.3 3-chlorobut-1-ene (2) 2.2.4 H H O H O H H C C O C H H C C O C H H H H H (2) Any one  (One mark for the functional group, one mark for the whole structure) [10] QUESTION 3 3.1 What is the relationship between the mass of a haloalkane and the boiling point? . (2) 3.2 boiling point  (1) 3.3 mass of the haloalkane / halogen  (1) 3.4 length of the carbon chain  (also accept: atmospheric pressure OR just pressure) (1) 3.5 So that the investigation may be deemed valid.  (also accept: Changes in pressure affect the boiling points of liquids) (1) 3.6 As the size / mass of the halogen attached to the methane increases, the boiling point increases. (2) 3.7 As the size / mass of the halogen attached to the methane increases, the size/mass of the whole compound increases, therefore the surface area increases and the strength of the intermolecular forces / Van der Waals forces / London forces, increases. More energy is needed to break the forces and therefore the boiling point increases. (3) [11] Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P2 4 September 2014 NSC-Memorandum QUESTION 4 4.1.1 I = addition / hydrogenation  II = addition / halogenation  III = substitution  IV = combustion (oxidation)  V = elimination / dehydrohalogenation  VI = polymerisation  (6) 4.1.2 I = H2  IV = O2  (2) 4.2.1 Esterification  (1) 4.2.2 H2SO4 is a dehydrating agent / catalyst. Will remove water (H2O) during the formation of the ester. Makes the reaction faster  (2) 4.2.3  Contents of the test tube must not be exposed to direct flame   Bumping stones reduce the intensity of the boiling action to prevent contents from shooting out.  (2) 4.2.4 H H H O H H H H C C C O + C C C H H H H  H O  H H O H H H H H C C C H + H O H H C C C O H H (5)  H H H  4.3.1 Ethene / ethylene (1) 4.3.2 Contains a double bond between two carbon atoms (1) 4.3.3 Prepare a solution of bromine water (brown). Add bromine water to a solution of ethane (ethane is a gas at room temperature). If the bromine water (2) decolourises quickly the presence of a double bond is indicated. Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P2 5 September 2014 NSC-Memorandum 4.3.4 H H H H H H (2) H C C C C C C H  H H H H H H 4.3.5 Addition  (1) [27] QUESTION 5 5.1 Increase the concentration of sulphuric acid  (4) Increase surface area of phosphate rock  5.2.1 The minimum kinetic energy of reacting particles  that can result in a reaction (2) 5.2.2 Collision theory  (1) 5.2.3 The shaded areas in the distribution curves represent the number of molecules with sufficient kinetic energy to overcome the activation energy . An increase in the temperature of the system results in a greater number of particles with sufficient kinetic energy to overcome the activation energy of the reaction . This results in more effective collisions per unit time OR a higher chance of an effective collision occurring, resulting in a higher reaction rate . (3) [10] Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P2 6 September 2014 NSC-Memorandum QUESTION 6 6.1.1 Dynamic (Equilibrium) – Refers to the fact that the forward and reverse reactions continue to take place simultaneously without stoppingand the rate of the forward reaction is equal to the rate of the reverse reaction, when (2) equilibrium is reached . (changes in macroscopic properties cease, changes in microscopic properties continue) 6.1.2 (a) 2SO2(g) + O2(g) → 2SO3(g)  (2) (b) 2SO3(g) → 2SO2(g) + O2(g)  (2) 6.1.3 There was an increase in pressure . This increases the rate of both the forward and reverse reactions due to the particles being closer together and effective collisions occurring more frequently. However, the forward reaction is favoured & therefore the rate of the forward reaction is increased more to turn 3 moles of gas into 2 moles of gas to relieve the increased pressure  (3) 6.1.4 The catalyst was removed  (1) 6.2 n(O2) = m M = 9,6 32 = 0,3 mol 2H2 (g) O2(g) 2H2O(g) Mol in the beginning 0,5 0,5 0 Mol reacted 0,4 0,2 0,4  ratio Mol at equilibrium 0,1 0,3  0,4  [ ] at equilibrium 0,5 1,5 2,0 divide by V(0,2) Kc = [H2O]2  [H2]2[O2] = (2,0)2 (0,5)2(1,5)  (7) = 10,67  [17] Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P2 7 September 2014 NSC-Memorandum QUESTION 7 7.1 pH = - log[H+]  = -log (3,2x10-5)  = 4,5  NO, the fish will not survive (4) 7.2 A substance that donates protons (H+) OR proton donor  (2) 7.3.1 C (Na2CO3)  (1) 7.3.2 A (HCℓ)  (1) 7.3.3 D (CH3COOH)  (1) [9] QUESTION 8 8.1.1 A monoprotic acid is an acid that can donate only one proton (H+-ion) per (2) molecule  8.1.2 H3O+ & CH3COO- (2) 8.1.3 HCℓ is a strong acid which ionises almost completely into its ions and yields a (2) high concentration of H3O+ ions in the solution, while ethanoic acid is only partially ionised and has less H3O+ ions in solution. 8.2.1 M(NaOH) = 40 g·mol-1 c = m MV = 2,5 (40)(0,5) = 0,125 mol·dm-3 (3) Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P2 8 September 2014 NSC-Memorandum 8.2.2 caVa = na cbVb nb ca = nacbVb nbVa = (1)(0,125)(20) (1)(24,5) = 0,102 mol·dm-3 (3) 8.2.3 The acid, When NaOH is allowed to stand for longer periods, it may evaporate (2) leaving solid sodium hydroxide and the tap will eventually not be able to turn. 8.2.4 Phenolphthalein, because a strong base is titrated with a weak acid and will form a basic (pH > 7) solution. The equivalence point of the indicator must (2) therefore be basic. [16] QUESTION 9 9.1 Ni  (1) 9.2 Ag+ + e- → Ag (2) 9.3 Eθcell = Eθcathode - Eθanode = 0,8 - (-0,27) = 1,07 V (4) 9.4 It acts as a salt bridge that separates the two electrolytes but enable ions to move from one half cell to the other to ensure electrical neutrality of the electrolyte solutions OR it completes the circuit  (3) 9.5 Mol ratio : Ni : e- OR 𝑛(𝑁𝑖) 1 = 1 : 2 𝑛(𝑒 − ) 2 𝑛(𝑁𝑖) 1 0,2 : 0,4  = 0,4 2 Therefore m=nxM 1 𝑛(𝑁𝑖) = × 0,4 2 = 0,2 x 59  = 0,2 = 11,8g  𝑚 = 𝑛𝑀 (4) = 0,2 × 59 = 11,8 𝑔 [14] Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P2 9 September 2014 NSC-Memorandum QUESTION 10 10.1 battery  e-  Anode: Copper electrode  Medal  Cu 2+ Cu 2+  CuSO4 or any other copper salt  (6) 10.2 Cu2++ 2e- → Cu  (2) 10.3 Cathode  (1) [9] QUESTION 11 11.1 26 % of 50 kg = 13 kg  Mass of N = 3 x 13  10 (3) = 3,9 kg  11.2 Ratio of N : K : P carefully controlled according to specific needs.any 1 (1) Nutrients are concentrated \ need less of them. 11.3 A = Nitrogen  B = Brine (seawater)  C = Haber Process  D = Water  E = Ostwald Process  Question asked for names don’t give formulae (5) [9] Copyright reserved Please turn over